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3 years ago

J2 + 7j + 12 = 0CAN U PLEASE HELP ME!!!

Mathematics

2 answers:

natima [27]3 years ago

3 0

Answer:

j = -3, -4

Step-by-step explanation:

j^2 + 7j + 12 = 0

(j + 3)(j + 4) = 0

j = -3, -4

allsm [11]3 years ago

3 0

Answer:

j=−3 or j=−4

Step-by-step explanation:

j2+7j+12=0

Step 1: Factor left side of equation.

(j+3)(j+4)=0

Step 2: Set factors equal to 0.

j+3=0 or j+4=0

j=−3 or j=−4

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3 years ago

Verify a(b-c)=ab-ac for a=1.6;b=1/-2;& c=-5/-7

harina [27]

Given:

$a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}$

To verify:

$a(b-c)=ab-ac$ for the given values.

Solution:

We have,

$a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}$

We need to verify $a(b-c)=ab-ac$ .

Taking left hand side, we get

$a(b-c)=1.6\left(\dfrac{1}{-2}-\dfrac{-5}{-7}\right)$

$a(b-c)=1.6\left(-\dfrac{1}{2}-\dfrac{5}{7}\right)$

Taking LCM, we get

$a(b-c)=1.6\left(\dfrac{-7-10}{14}\right)$

$a(b-c)=\dfrac{16}{10}\left(\dfrac{-17}{14}\right)$

$a(b-c)=\dfrac{8}{5}\left(\dfrac{-17}{14}\right)$

$a(b-c)=-\dfrac{68}{35}\right)$

Taking right hand side, we get

$ab-ac=1.6\times \dfrac{1}{-2}-1.6\times \dfrac{-5}{-7}$

$ab-ac=-\dfrac{16}{20}-\dfrac{8}{7}$

$ab-ac=-\dfrac{4}{5}-\dfrac{8}{7}$

Taking LCM, we get

$ab-ac=\dfrac{-28-40}{35}$

$ab-ac=\dfrac{-68}{35}$

Now,

$LHS=RHS$

Hence proved.

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2 years ago

J2 + 7j + 12 = 0CAN U PLEASE HELP ME!!! ​

J2 + 7j + 12 = 0CAN U PLEASE HELP ME!!!